1. The problem is to evaluate the double sum $$\sum_{i=0}^3 \sum_{j=1}^4 (i^2 + j)$$. 2. The formula for a double sum is to sum over the inner index first, then the outer index. 3. We start by fixing $i$ and summing over $j$: $$\sum_{j=1}^4 (i^2 + j) = \sum_{j=1}^4 i^2 + \sum_{j=1}^4 j = 4i^2 + (1+2+3+4)$$ 4. Calculate the sum of $j$ from 1 to 4: $$1+2+3+4 = 10$$ 5. So the inner sum becomes: $$4i^2 + 10$$ 6. Now sum over $i$ from 0 to 3: $$\sum_{i=0}^3 (4i^2 + 10) = 4\sum_{i=0}^3 i^2 + \sum_{i=0}^3 10 = 4(0^2 + 1^2 + 2^2 + 3^2) + 10 \times 4$$ 7. Calculate the sum of squares: $$0 + 1 + 4 + 9 = 14$$ 8. Substitute back: $$4 \times 14 + 40 = 56 + 40 = 96$$ 9. Therefore, the value of the double sum is $$\boxed{96}$$.